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exam 4 reviews - By Joey and Kaitlin Yields Theoretical...

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Unformatted text preview: By Joey and Kaitlin Yields Theoretical Yield- Is the maximum amount of a given product that can be formed when the limiting reagent is completely consumed Experimental Yield- Is the amount of product actually obtained from the chemical reaction Percent Yield- Is the actual yield of a product as the percentage of the theoretical yield How to Calculate % Yield= Experimental Yield/Theoretical Yield 1 st determine the limiting reagent (the one that runs out first) The limiting reagent determines how much product is possible to be obtained which is the theoretical yield Example 1-determine limiting reagent CH4 + 2 O2 CO2 + 2 H2O If your reaction contains 0.50 moles of CH4 and 0.60 mol O2 which one is the limiting reagent? 0.50 mol CH4 x 1 mol CO2 = 0.50 mol of CO2 1 mol of CH4 0.60 mol O2 x 1 mol CO2 = 0.30 mol CO2 2 mol O2 Example 2-Theoretical Yield SiO2 + 4HF SiF4 + 2H2O How many grams of SiF4 can be produced from 78.9 g of SiO2 with an excess of HF? SiO2 = 60.01 g/mol SiF4 = 104.01 g/mol 78.9 g SiO2 x 1 mol SiO2 x 1 mol SiF4 x 104.01 g SiF4 = 137 g SiF4 60.01 g/mol SiO2 1 mol SiO2 1 mol SiF4 Example 3-Percent Yield Say that you only managed to produce 105.6 g of SiF4 from the reaction from example 2 what would your percent yield be? 105.6 g SiF4 x 100 = 77.1 % 137 g SiF4 Example 4 2 C2H 6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 45 g of C2H 6 and 30 g of O2 when only 18.5 g of CO2 are actually produced. C2H 6 =30.07 g/mol O2 = 32.0 g/mol CO2 = 44.01 g/mol 45.0 g C2H 6 x 1 mol C2H6 x 4 mol CO2 x 44.01 g/mol CO2 = 131.7 g CO2 30.07g/molC2H6 2mol C2H6 1 mol CO2 30.0 g O2 x 1 mol O2 x 4 mol CO2 x 44.01 g/mol CO2 = 23.6 g CO2 32.0 g/molO2 7 mol O2 1 mol CO2 Step One: Limiting Reagent Example 4 2 C2H6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 45 g of C2H6 and 30 g of O2 when only 18.5 g of CO2 are actually produced. C2H6 =30.07 g/mol O2 = 32.0 g/mol CO2 = 44.011 g/m Experimental Yield 18.5 g CO2 x 100 = 78.4 % Theoretical Yield 23.6 g CO2 30.0 g O2 x 1 mol O2 x 4 mol CO2 x 44.011 g/mol CO2 = 23.6 g CO2 32.0 g/molO2 7 mol O2 1 mol CO2 Step Two: Theoretical Yield You just found it! 23.6 g of CO2 can be produced from 30 g of O2. Step Three: Percent Yield Example 5 2 C2H 6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 18.0 ml of C2H 6 with excess O2 and 4.06 x 10-2 g of CO2 are actually produced. C2H 6 = 30.07 g/mol C2H 6 (density)= 1.34 x 10-3 g/ml CO2 = 44.01 g/mol 18.0 ml C2H6 x 1.34 x 10- 3 g x 1 mol C2H 6 x 4 mol CO2 x 44.011 g CO2 = 7.06 x 10-2 g ml C2H 6 30.07 g 2 mol C2H 6 1 mol CO2 4.06 x 10-2 g C2H 6 x 100 = 57.5 % 7.06 x 10-2 g C2H 6 Example 6 If 6.3 g of H2O is produced in the following Combustion reaction C 3 H 8 + 5O 2 4H 2 O + 3CO 2 containing 14 g of Propane ( C3H 8 ) and 14 g of O2, How many grams will I have left over of O 2 and Propane?...
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This note was uploaded on 02/05/2011 for the course CHEM 151 taught by Professor Staff during the Fall '08 term at Arizona.

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exam 4 reviews - By Joey and Kaitlin Yields Theoretical...

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